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Readme [11.4K]
3 years ago
7

What is the solution of 7 + m ≥ 2 and m + 1 < 2?

Mathematics
1 answer:
dolphi86 [110]3 years ago
5 0

Answer:

-5 ≤m<1

Step-by-step explanation:

7 + m ≥ 2 and m + 1 < 2?

Solve the inequality on the left first

7 + m ≥ 2

Subtract 7 on each side

7-7 + m ≥ 2-7

m  ≥ -5


Now solve the equation on the right

m + 1 < 2

Subtract 1 from each side

m + 1-1 < 2-1

m <1

Since this is an and

-5 ≤m<1

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Solve: (-5/6) × (9/20) + (-5/6) × 7/25 = ?​
jonny [76]

\bf \underline{★ How \:to\: do -} \\

Here, we are given with four fractions to multiply two of them and to add two of them. If we add them directly by taking the LCM and adding them is not a similar way. We can clearly observe that in those four fractions, we have two fractions as common i.e, we have two fractions as same. If we have two fractions or numbers as same, we can solve the sum by an other concept called as distributive property. In this property, we multiply the common fraction with the sum of other two fractions. This concept can also be done with fractions as well as integers. So, let's solve!!

\:

\bf \underline{➤ Solution-} \\

{\tt \leadsto \dfrac{(-5)}{6} \times \dfrac{9}{20} + \dfrac{(-5)}{6} + \dfrac{7}{25}}

Group the non-common fractions in bracket.

{\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{9}{20} + \dfrac{7}{25} \bigg)}

First we should solve the numbers in bracket.

LCM of 20 and 25 is 100.

{\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{9 \times 5}{20 \times 5} + \dfrac{7 \times 4}{25 \times 4} \bigg)}

Multiply the numerators and denominators in the bracket.

{\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{45}{100} + \dfrac{28}{100} \bigg)}

Now, write both numerators in bracket with a common denominator.

{\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{45 + 28}{100} \bigg)}

Now, add the numerators in bracket.

{\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{73}{100} \bigg)}

Write the numerator and denominator in lowest form by cancellation method.

{\tt \leadsto \dfrac{\cancel{(-5)} \times 73}{6 \times \cancel{100}} = \dfrac{(-1) \times 73}{6 \times 20}}

Now, multiply the numerators and denominators.

{\tt \leadsto \dfrac{(-73)}{120}}

\:

{\red{\underline{\boxed{\bf So, \: the \: answer \: obtained \: is \: \: \dfrac{(-73)}{120}}}}}

3 0
3 years ago
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